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Registered Member #1617
Joined: Fri Aug 01 2008, 07:31AM
Location: Adelaide, South Australia
Posts: 139
What is an easy way to switch a high side mosfet, using discrete components (no high side gate drive IC) in a H-bridge circuit. I've seen mentioned in various places 'bootstrap' and charge pump type circuits, but i cant find any clear examples of how to implement these, or any real explaination as to how either would work. Can someone explain the basics of what these look like/how they work to me?
Registered Member #1617
Joined: Fri Aug 01 2008, 07:31AM
Location: Adelaide, South Australia
Posts: 139
when i click the link for explaination, i get a page telling me access denied. Can some one explain how this circuit works?
Or with optocouplers
how could it be done with optos? from what i understand, the main issue is getting the higher voltage, the optos provide the isolation, but i would like to avoid having a seperate floating supply?
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
The link is to a thread on the old 4hv that was archived.
This is the picture on that page: it's about the simplest high-side driver you can possibly make. It actually uses the low-side power MOSFET as the level shifter for the high-side gate drive, as well as a power switch in its own right. As far as I know, I invented this topology, unfortunately it's of no commercial use because D1 has to carry the full load current and withstand the DC bus voltage, which makes it a large, expensive, lossy component.
High-side gate drive ICs work in a vaguely similar way, just integrated on one chip.
Most of them can't keep the high-side switch on indefinitely. It must be turned off every so often, and the low-side turned on, to recharge the bootstrap capacitor (which is C1 in my circuit)
You also get optoisolated gate drive ICs like the TLP250.
Registered Member #1617
Joined: Fri Aug 01 2008, 07:31AM
Location: Adelaide, South Australia
Posts: 139
thanks steve, but im still unclear as to how the 'bootstrap' concepts works. so when the low side switch is on, this charges the cap via the diode to twice the DC bus voltage, referenced to the source of the upper switch? so you have essentially a floating 'voltage source' (the bootstrap capacitor) referenced to the source of the upper switch? is this aproximately what happens? What does D1 actually do?
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Think "pulling yourself up by your own bootstraps". The bootstrap capacitor is a portable source of gate voltage that the high-side MOSFET carries with it, by virtue of its negative terminal being connected to the FET source.
So say the low-side FET is on. The bootstrap cap charges to 12V from the 12V supply through D2, R4.
Now let's turn the low-side FET off and the high-side on. The bootstrap capacitor is now released and charges the high-side FET (M2) gate to 12V through R3. M2 begins to turn on, and its source voltage starts to rise from zero towards the DC bus. As it does so, it carries the bootstrap capacitor with it, so the gate is still driven with 12V relative to the source. Therefore it will continue turning on until the drain and source are both at the DC bus voltage, and the gate is at DC bus voltage + 12. Whoops, your free tech support just ran out!
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High side switch
